Uniform rod pivoted off-center

[natasha13100]

√ω

Homework Statement

A uniform rod of mass 1.2 kg and length 1.2 m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released. The acceleration due to gravity is g = 9.8 m/s².
Part 1. Find the rotational inertia of the rod relative to the axis perpendicular to the screen and passing through the pivot point (in kgm²).
Part 2. As the rod swings downward, its angular speed velocity and the magnitude of its angular acceleration, respectively are (increasing, decreasing, staying the same) and (increasing, decreasing, staying the same).
Part 3. What is the angular speed (in rad/s) of the rod as it passes through the vertical position (when end marked B is at the bottom)?
Part 4. What is the linear speed of the bottom end of the rod (marked B) when it passes the vertical position?(in m/s)

Homework Equations

I_cm=1/12ml²
I_p=I_cm+md²
U_g=mgh
U_i+K_i+W^noncon=U_f+K_f
K_trans=1/2mv²
K_rot=1/2Iω²
t=Frsinθ
K=K_trans+K_rot
v=ωR

The Attempt at a Solution

Picture/FBD is attached.
Part 1
I_p=1/12m(l)²+m(l/4)²=1/12*1.2*1.2²+1.2*.3²=.252
Part 2
increasing and decreasing
Part 3
0=1/2mv²+1/2Iω²+mgh where h=-l/4(distance from pivot to cm which will be directly below pivot when the rod is at the bottom of its swing)
mgh=1/2mv²+1/2Iω²=mgh=1/2mω²R²+1/2Iω²=1/2mω²R²+1/2Iω²
1.2(9.8)(.3)=(1/2(1.2)(.3)²+1/2*.252)ω²=(.18)ω²
ω=√(3.528/.18)=4.42718872424
I am not sure if this is correct, but I used l/4 (distance from pivot to center of mass) as R.
Part 4
mgh=1/2mv²+1/2Iω²
2mgh=mv²+Iv²/R²
2(1.2)(9.8)(.3)=(1.2+.252/.3²)v²
v=√(7.056/4)=1.32815661727
or v=ωR²=4.42718872424(.3)^2=0.398446985181
Since these two values for v don't match up, I think I'm doing something wrong.

 

[tiny-tim]

hi natasha13100!

i think you're using the wrong formula for kinetic energy …

it's 1/2mv² of the c.o.m. plus 1/2I_c.o.mω² about the c.o.m.

(which also equals 1/2I_c.o.rω² about the instantaneous centre of rotation, ie the pivot)

 

[natasha13100]

What does that mean?
So K=1/2mv^2+1/2(1/12ml^2)w^2=1/2mw^2/(l/2)^2+1/24ml^2w^2=w^2(ml^2/8+ml^2/24)
Uf+Kf=Ui+Ki
0=w^2(ml^2/8+ml^2/24)-mgl/4
mgl/4=w^2(ml^2/8+ml^2/24)
g=w^2(l/2+l/6)
w^2=g/(2l/3)
w=sqrt(3g/2l)
and
v=wl/2=sqrt(3g/2l)l/2
or
K=1/2mv^2+1/2(1/12ml^2)w^2=1/2mv^2+1/2(1/12ml^2)v^2/(l/2)^2=v^2(ml/2+1/6m)
gl/4=v^2(l/2+1/6)
v^2(3l+1)/6=gl/4
v^2=3/(6l+2)
v=sqrt(3/(6l+2))
but v is still off

 

[tiny-tim]

the v in 1/2 mv² will be (l/4)ω

(because l24 is the distance from the centre of rotation to the centre of mass)

 

[natasha13100]

0=w^2(1/2ml^2/16+ml^2/24)-mgl/4
mgl/4=w^2(ml^2/32+ml^2/24)
g=w^2(l/8+l/6)
w^2=24g/(3l+4l)
w=sqrt(24g/7l) which is correct
and
v=lsqrt(24g/7l)/4 but that's not right.

 

[tiny-tim]

0=w^2(1/2ml^2/16+ml^2/24)-mgl/4
mgl/4=w^2(ml^2/32+ml^2/24)
g=w^2(l/8+l/6)
w^2=24g/(3l+4l)
w=sqrt(24g/7l)

looks ok down to here

(btw, you'd get the same result if you used the parallel axis theorem, and just 1/2 I_c.o.r.ω² )

 

[natasha13100]

1/2I_c.o.r.ω²=mgl/4
_c.o.r.=I_c.o.m.+m(l/4)²
I_c.o.m.=1/12ml²
1/2(1/12ml²+m(l/4)²)ω²=mgl/4
ω²=g/(l(1/6+1/8))=g/l/(7/24)
ω=√(24g/7l)
okay, I see what you mean.
For v, wouldn't you just plug in for ω?
1/2mv²+1/2(1/12ml²)ω²=mgl/4
m/2*v²=mgl/4-1/24ml^2*24g/7l=mgl/4-mgl/7
v²=(7gl-4gl)/14=3gl/14
v=√(3gl/14)

 

[tiny-tim]

For v, wouldn't you just plug in for ω?

yes

(but in your last post you used the wrong distance … the question asks for v for the bottom of the rod, which is not the same v as you used earlier)

1/2mv²+1/2(1/12ml²)ω²=mgl/4
m/2*v²=mgl/4-1/24ml^2*24g/7l=mgl/4-mgl/7
v²=(7gl-4gl)/14=3gl/14
v=√(3gl/14)

i'm confused … what is this?

 

[natasha13100]

Oh that was the v of the center of mass (I think). So for the bottom of the rod, r=3l/4 (distance from pivot to bottom of rod) and v=√(24g/7l)*3l/4

 

[tiny-tim]

yup!